[Rutherford]

Thallium exists in two different oxidation states: Tl⁺ and Tl³⁺. The standard redox potentials for some relevant reactions are:
Tl⁺(aq) + e– → Tl(s) Eº₁ = – 0.336 V
Tl³⁺(aq) + 3e– → Tl(s) Eº₂ = + 0.728 V
Calculate the redox potential Eº₃ for the following reactions: Tl³⁺(aq) + 2 e– → Tl⁺(aq).

I converted those potentials to equilibrium constants and K=K₂/K₁ then I converted this constant back to potential and I got 1.26V which is the answer. Why don't I get the same result when I deduct Eº₁ from Eº₂ (like when I am using enthalpies)?

[XGen]

I don't think reduction potential works like that just by virtue of how it is.

You can do a couple things to solve for the new potential. One is like what you did, and another is using Gibb's free energy calculations.

[Schrödinger]

Exactly. ΔG° = nFE°
The number of moles of electrons being transferred is not the same in these reactions. So, the E°'s are not additive. It's always better to convert the E° values to ΔG° values and do the usual algebra. That way, it is less confusing and less error-prone

[Rutherford]

To prove this to myself I actually did some maths. As the potential at the equilibrium is 0, the standard potentials are equal:
Eº₁=RT*lnK₁/(z₁F) (I usually mark the number of electrons with z)
Eº₂=RT*lnK₂/(z₂F)
Eº₂-Eº₁=RT*ln(K₂^z₁/K₁^z₂):(z₁z₂F)
The new equilibrium constant should be K₂/K₁ and not K₂^z₁/K₁^z₂, so really, the number of electrons makes it not additive. Thanks to you two for the help. I usually do this by converting the potentials to eq. constants, but converting it to Gibb's energy and do it like for the enthalpies is good, too.

[Schrödinger]

Yeah. The reason I referred to the Gibbs' energy pathway is because we're used to adding thermodynamic heats/enthalpies/energies and that will not cause any confusion, since they are expressed in J/mol or kJ/mol. Hence you know that you need to multiply the quantity by number of moles to get the total heat/enthalpy/energy